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math-ph/0004014
Suppose MATH is in the MATH-class and define MATH as in REF . Suppose MATH (for a proof of this statement, see REF ). The argument hinges on particular scaling properties of the functionals MATH and MATH, which enable us to convert REF into REF. Given MATH, let us for each MATH define MATH by MATH . Then we have MATH where in the first relation we used that MATH can be converted into MATH in REF by substituting MATH in the place of MATH; the second relation is a result of a simple spatial scaling of the first line in REF. Note that MATH. Let MATH be a minimizing sequence of the variational problem in REF. Suppose, without loss of generality, that MATH and MATH. Then we have MATH . Now pick any MATH and consider instead the sequence MATH. Clearly, MATH for all MATH. By REF, the derivative of the right-hand side must vanish at MATH, that is, MATH . By putting REF together, we easily compute that MATH . Note that while MATH is strictly increasing, MATH is strictly decreasing. This allows us to recast REF as MATH . Indeed, we begin by observing that MATH holds in REF, as is verified by pulling MATH inside the bracket, replacing it with MATH, and dropping the last condition. To prove the MATH part, note that the above sequence MATH for MATH eventually fulfills the last condition in REF because MATH. Since MATH, the right-hand side of REF is no more than MATH for any MATH. Taking MATH and recalling REF proves the equality in REF. With REF in the hand we can finally prove REF. By using MATH instead of MATH in REF, the condition MATH becomes MATH and the factor MATH appears in front of the infimum. Thus, setting MATH, which by REF requires that MATH (note that MATH) and invoking REF, we recover the variational REF . Therefore, MATH . From this, REF follows by simple algebraic manipulations. The claim MATH is a consequence of REF and the fact that MATH.
math-ph/0004014
REF for MATH is well-known. Assume that MATH and observe that, due to the perfect scaling properties of both MATH and MATH, REF can alternatively be written as MATH . Let MATH be the principal eigenvalue respectively, an associated eigenvector of MATH in MATH with NAME boundary condition. Then MATH, which means that MATH . Since MATH is continuous and bounded, the integral is finite, whereby MATH. REF and the remainder of REF are then simple consequences of the following observation, whose justification we defer to the end of this proof: MATH where MATH is the volume of the unit sphere in MATH. Indeed, to get that MATH is non-vanishing, set MATH and choose MATH such that the infimum in REF is strictly larger than MATH for all MATH. Clearly, MATH is finite, so MATH. Then for any MATH either MATH, which implies MATH, or MATH which implies MATH . Thus, in both cases, MATH independent of MATH. Since MATH is decreasing, the restriction to MATH is irrelevant which finishes REF . To prove also REF , note first that MATH for all MATH. Given MATH, let MATH be such that the infimum in REF is larger than MATH in REF. Consider REF restricted to MATH with MATH. Since for any such MATH the restricted infimum is no less than MATH. Therefore, MATH, which by MATH and REF proves REF. It remains to prove REF. To that end, denote the infimum by MATH and note that MATH . Indeed, denoting MATH for any MATH, we have MATH, MATH, and MATH, whereby REF immediately follows. Since MATH, it suffices to prove REF just for MATH. Recall that the operator MATH on MATH with NAME boundary condition has a compact resolvent, so its spectrum MATH is a discrete set of finitely-degenerate eigenvalues. For each MATH, define the function MATH . Then MATH, with MATH and the eigenvectors given as MATH. Note that the latter form a REF basis in MATH. Let MATH and MATH be fixed. Let MATH be such that MATH. Note that MATH. Pick a function MATH such that MATH and let MATH. Let MATH respectively, MATH be the normalized projections of MATH onto the NAME spaces generated by MATH with MATH respectively, MATH. Then MATH with MATH. We claim that MATH. Indeed, MATH where MATH is such that MATH for all MATH with MATH and MATH . Here we used that MATH, then we applied NAME inequality and noted that MATH is normalized to one in MATH, because MATH for all MATH. The third inequality follows by the observation MATH implied by MATH. Let MATH be such that MATH. Then we have MATH. Using REF, we derive that MATH, whereby we have that MATH. This gives us the bound MATH that is, MATH. On the other hand, MATH where we used that MATH and that MATH has no overlap with MATH such that MATH. By putting REF together and noting that MATH, REF for MATH follows.
math-ph/0004014
Let MATH and consider MATH. Let MATH and define MATH. By a straightforward calculation, MATH. Let MATH. Then MATH, while MATH. This implies that MATH. Since MATH, the proof is finished.
math-ph/0004014
In the case MATH, use the shift-invariance of MATH, NAME 's inequality, and the monotonicity REF to obtain MATH . In the case MATH, instead of NAME 's inequality we apply MATH to deduce similarly as in REF that MATH .
math-ph/0004014
By combining REF and the left inequality in REF for MATH instead of MATH, we see that MATH. Since MATH, the left-hand side of REF, with MATH instead of ``MATH," is bounded below by MATH. By REF , MATH positive, finite and non-zero.
math-ph/0004014
Recall the notation of Subsection REF. By taking the expectation over MATH (and using that MATH is an i.i.d. field) and recalling REF, we have for any MATH that MATH . Consider the scaled version MATH of the local times MATH . Let MATH be the space of all non-negative NAME almost everywhere continuous functions in MATH with a bounded support. Clearly, MATH and MATH. Introduce the functional MATH, assigning each MATH the value MATH where we recalled REF. Substituting MATH and MATH into REF, we obtain MATH . Using shift-invariance and the fact that MATH for any MATH, we have MATH . It is well known that the family of scaled local times MATH satisfies a weak large-deviation principle on MATH with rate MATH and rate function MATH defined in REF. This fact has been first derived by CITE for the discrete-time random walk; for the changes of the proof in the continuous time case we refer to REF of the monograph by CITE. The large-deviation principle allows us to use NAME 's integral lemma to convert both bounds in REF into corresponding variational formulas. Note that, if both MATH and MATH are appropriately extended to MATH, all infima REF can be taken over MATH with the same result. In the sequel, we have to make a distinction between the cases MATH and MATH. In the case MATH, our Scaling Assumption implies that, for every MATH, MATH is continuous and MATH converges to MATH uniformly on the space of all measurable functions MATH with MATH topology. Indeed, for any such function MATH and any MATH, the integral REF can be split into MATH and MATH. The former then converges uniformly to MATH, while the latter can be bounded as MATH where we invoked the monotonicity of MATH. Taking MATH proves that this part is negligible for MATH and, if MATH is invoked before MATH, it also shows that MATH uniformly in MATH as MATH. Having verified continuity, NAME 's lemma (and MATH) readily outputs the left inequality in REF, while on the right-hand side it yields a bound in terms of the quantity MATH defined in REF. By REF , MATH tends to MATH as MATH, which proves the inequality on the right of REF. In the case MATH, the lower bound goes along the same line, but we have to be more careful with REF, since MATH in this case. Let us estimate MATH where we invoked the explicit form of MATH. Since both absolute values on the right-hand side tend to MATH as MATH uniformly in MATH, the lower bound in REF follows again by NAME 's lemma and limit MATH. For the upper bound, the estimate and uniform limit MATH give us a bound in terms of the quantity MATH defined in REF. By then MATH is irrelevant, so by invoking REF , the claim is proved by taking MATH. It remains to prove REF. Recall the shorthand MATH. By REF and analogously to REF, we have MATH . Noting that MATH, we thus have MATH. With this in the hand, REF directly follows by the right inequality in REF.
math-ph/0004014
In the course of the proof, we use abbreviations MATH and MATH. Recall that MATH denotes an orthonormal basis in MATH (with inner product MATH) consisting of the eigenfunctions of MATH with NAME boundary condition. We first turn to the case MATH. Use the NAME expansion REF and the inequality MATH to obtain MATH . By NAME 's inequality for the probability measure MATH we have MATH where we recalled from the end of the proof of REF that MATH, inserted MATH, and applied REF. In the case MATH, we apply NAME 's inequality as follows: MATH . Invoking that MATH, the proof is finished by recalling REF once again.
math-ph/0004014
Let MATH. First, notice that the second term in REF can be estimated in terms of a sum: MATH . Thus, applying REF to MATH (that is, for MATH) with MATH replaced by MATH for some fixed MATH, raising both sides to the MATH-th power, and using REF we get MATH . Next we take the expectation with respect to MATH and note that, by the shift-invariance of MATH, the distribution of MATH does not depend on MATH. Take logarithm, multiply by MATH and let MATH. Then we have that MATH where we also used that MATH, MATH, and MATH are all MATH as MATH. Since MATH REF for MATH instead of MATH implies that the second term on the right-hand side of REF is bounded by MATH. The upper bound in REF then follows by letting MATH.
math-ph/0004014
It is immediate from REF with MATH that MATH . According to REF, we have, for every MATH, MATH . Using this for MATH in REF, we see that, for sufficiently large MATH (depending only on MATH and MATH), the right-hand side of REF is no more than MATH.
math-ph/0004014
The idea is to construct a partition of unity MATH where MATH with MATH . Then we put MATH where MATH is the discrete gradient. Obviously, MATH is MATH-periodic in every component. The construction of MATH such that MATH satisfies REF is given at the end of this proof. Assuming the existence of the above partition of unity, we turn to the proof of REF. Recall the NAME REF , which can be shortened as MATH, where MATH and where the supremum is over normalized MATH with support in MATH. Let MATH be such a function, and define MATH for MATH. Note that, according to REF, we have MATH and MATH. The pivotal point of the proof is the bound MATH . In order to prove this inequality, we invoke the rewrite MATH recall REF, and then perform a couple of symmetrizations to derive MATH where MATH is given by the formula MATH . Using this bound on the right-hand side of REF, we have MATH which is exactly REF . Since the support of MATH is contained in MATH, the NAME formula yields that MATH whenever MATH (which requires, in particular, that MATH). Estimating these eigenvalues by their maximum and taking into account that MATH, we find that the right-hand side of REF does not exceed the right-hand side of REF . REF is finished by passing to the supremum over MATH on the left-hand side of REF. For the proof to be complete, it remains to construct the functions MATH and MATH with REF and such that MATH for some MATH. First, the ansatz MATH reduces the construction of MATH to the case MATH (with MATH replaced by MATH). In order to define MATH, let MATH be such that both MATH and MATH are smooth, MATH on MATH and MATH on MATH and MATH for all MATH. Then we put MATH . In order to verify that the functions MATH with MATH form a partition of unity on MATH, we first note that MATH on MATH while MATH for MATH. Moreover, as follows by a direct computation, MATH, which means that REF is satisfied with MATH. This finishes the construction and also the proof.
math-ph/0004014
Having all the prerequisites, the proof is easily completed. First, MATH by REF . Therefore, combining REF with REF , we have that MATH whenever MATH is large enough. Invoking also the NAME expansion REF with respect to the eigenfunctions of MATH in MATH and the fact that MATH, we find that MATH . Now apply REF for MATH to finish the proof.
math-ph/0004014
From REF we immediately have MATH where MATH is as in REF. Here, for the upper bound we simply neglected MATH in REF, whereas for the lower bound we first wrote REF as a normalized sum of the right-hand side of REF with the walk starting and ending at all possible MATH, and then inserted MATH, applied REF, and then recalled REF. The factor MATH comes from the normalization by MATH in the first step. Using subsequently REF for MATH, the left-hand side of REF is further bounded from below by MATH. Then REF and the limit MATH enable us to conclude that MATH . In the remainder of the proof, we have to convert this statement into the appropriate limit for the IDS. This is a standard problem in the theory of NAME transforms and, indeed, there are theorems that can after some work be applied (for example, NAME 's NAME Theorem, see CITE). However, for the sake of both completeness and convenience we provide an independent proof below. Suppose that MATH is the MATH-class. We begin with an upper bound. Clearly, MATH . Let MATH and insert this for MATH in the previous expression. The result is MATH where we applied REF and the definition of MATH. In order to finish the upper bound, we first remark that from the first assertion in REF it can be deduced that MATH . Indeed, define MATH and consider the quantity MATH. Clearly, MATH . Let MATH. Since MATH as MATH, there is no MATH such that MATH for infinitely many MATH with an accumulation point at zero, because otherwise the left-hand side REF would, by REF, eventually exceed the right-hand side. Similarly we prove that MATH cannot be smaller than MATH. Therefore, MATH as MATH, which is REF. Using REF, we have from REF that MATH . The lower bound is slightly harder, but quite standard. First, introduce the probability measure on MATH defined by MATH . We claim that, for any MATH, all mass of MATH gets eventually concentrated inside the interval MATH as MATH. Indeed, for any MATH we have MATH . Pick MATH and set MATH. Then we have MATH where we again used REF. Applying that MATH, using MATH and noting that MATH, we have MATH . Choosing MATH small enough, the right-hand side vanishes as MATH. Similarly we proceed in the case MATH. Now we can finish the lower bound on NAME tails. Indeed, using NAME 's inequality MATH . But MATH tends to MATH, by what we have proved about the concentration of the mass of MATH (note that REF and the similar bound for MATH are both exponential in MATH) and, by the same token, so does MATH. By putting all this together, dividing both sides of REF by MATH with MATH, interpreting MATH as a new variable tending to MATH as MATH, and invoking REF and the subsequent computation, we get MATH where we also used that MATH. Since MATH was arbitrary, the claim is finished by taking MATH.
math-ph/0004014
Let MATH and let MATH. We want to apply REF with the random potential MATH and with MATH replaced by MATH for some fixed MATH. (Later we shall let MATH and pick MATH appropriately.) Recall the definition of MATH in REF and abbreviate MATH. Take logarithms in REF, multiply by MATH and use REF to obtain MATH almost surely with respect to the field MATH. Thus, we just need to evaluate the almost sure behavior of the maximum of the random variables on the right-hand side. This will be done by showing that MATH almost surely with respect to the field MATH, provided MATH is chosen appropriately. For any MATH, let MATH be an enumeration of the random variables MATH with MATH. Note that MATH for MATH large. Clearly, MATH are identically distributed but not independent. By REF, the tail of their distribution is bounded by MATH where MATH is defined in REF. REF will be proved if we can verify that, with probability one, MATH for any MATH and sufficiently large MATH, as MATH. To that end, note first that the left-hand side of REF is increasing in MATH since the maps MATH, MATH and MATH are all increasing. As a consequence, it suffices to prove REF only for MATH, because also MATH as MATH for any MATH. Let MATH . Abbreviating MATH and recalling MATH, the exponential NAME inequality and REF allow us to write for any MATH and MATH large that MATH . Now let MATH to minimize the function MATH on MATH. An easy calculation reveals that MATH. By invoking REF , we also find that MATH for this value of MATH, and, substituting this into REF, we obtain MATH which is clearly summable on MATH provided MATH is sufficiently large. The NAME lemma then guarantees the validity of REF, which in turn proves REF. The limit MATH then yields the upper bound in REF .
math-ph/0004014
Let MATH and let MATH be twice continuously differentiable with MATH. If MATH, let MATH be a non-degenerate ball in MATH centered at MATH. Suppose that MATH does not belong to the exceptional null sets of the preceding assertions. In particular, there are unique infinite clusters MATH in MATH and MATH in MATH, and MATH satisfies the claims in REF . Clearly, MATH. Assume MATH and pick a MATH. For each MATH choose a MATH such that REF holds. We assume that MATH is so large that MATH. The lower bound on MATH will be obtained by restricting the random walk MATH (which starts at REF) to be at MATH at time MATH, at MATH at time MATH (staying within MATH in the meantime) and to remain in MATH until time MATH. Introduce the exit times from MATH and MATH, respectively, MATH . Let MATH. Inserting the indicator on the event described above and using the NAME property twice at times MATH and MATH, we get MATH where the three factors are given by MATH . Clearly, the quantity MATH is independent of MATH and is non-vanishing because MATH. Our next claim is that MATH as MATH. Indeed, MATH since there is at least one path connecting MATH to MATH within MATH (recall that the field MATH is bounded from below by MATH on MATH). Denote by MATH the minimal length of such a path and abbreviate MATH, where MATH is as in REF. Then, for MATH, MATH by REF and the fact that the both MATH. Hence, using also that MATH, MATH . In order to see that MATH, recall that MATH as MATH by REF and that MATH does not depend on MATH. We turn to the estimate of MATH. By spatial homogeniety of the random walk, we have MATH where MATH is the first exit time from MATH. Using REF, we obtain the estimate MATH . By invoking REF, the expectation on the right-hand side is bounded from below by MATH where MATH respectively, MATH denote the principal NAME eigenvalue respectively, the MATH-normalized principal eigenfunction of MATH in MATH. For MATH and MATH we have the following bounds, whose proofs will be given subsequently: We have MATH . Summarizing all the preceding estimates and applying REF, we obtain MATH where we also noted that MATH. In the case MATH, let MATH, optimize over MATH with MATH (clearly, the supremum in REF may be restricted to the set of twice continuously differentiable functions MATH such that MATH) and let MATH to get the lower bound in REF . In the case MATH, recall that MATH. It is classical (see, for example, REF , or argue directly by NAME 's inequality) that the supremum REF can be restricted to MATH whose support is a ball. The proof is therefore finished by letting MATH, optimizing over such MATH and letting MATH.
math-ph/0004014
We begin with REF. Recall that MATH is also an eigenfunction for the transition densities of the random walk in MATH with potential MATH. Using this observation at time MATH, we can write MATH . Since MATH is nonpositive and MATH is bounded from below, we have MATH . Using the same strategy as in REF, we have MATH. Since MATH is nonnegative and satisfies MATH we have MATH. From these estimates, REF is proved by noting that MATH. In order to establish REF, we shall restrict the supremum in REF to a particular choice of MATH. Let MATH if MATH and MATH if MATH. Let MATH be the MATH-normalized principal eigenfunction of the (continuous) operator MATH on MATH with NAME boundary conditions. Let us insert MATH into REF in the place of MATH. Thus we get MATH where MATH denotes that MATH and MATH are nearest neighbors. Since MATH is smooth, standard theorems guarantee that MATH is continuously differentiable on MATH and, hence, MATH. (This fact is derived using regularity properties of NAME 's function of the NAME equation, see, for example, REF.) Then MATH where MATH for some MATH. For the pairs MATH with MATH we only get a bound MATH (note that MATH in this case). Since the total contribution of these boundary terms to REF is clearly bounded by MATH, we see that the right-hand side of REF converges to MATH as MATH. By our choice of MATH, this limit is equal to the eigenvalue MATH, which proves REF.
math-ph/0004014
Suppose that MATH. This implies that MATH almost surely and, by the law of large numbers, MATH . Suppose that MATH does not belong to the exceptional sets of REF . For sufficiently large MATH, let MATH be such that REF holds. Let MATH. The strategy for the lower bound on MATH is that the random walk performs MATH steps toward MATH, resting at most time MATH at each site MATH between MATH and MATH, so that MATH is reached before time MATH. Afterwards the walk stays at MATH until MATH. Use MATH to denote the latter event. Then MATH, where MATH is as in REF and MATH. The lower bound on MATH is identical to the case MATH. To estimate the term MATH, suppose that MATH (clearly, if MATH no estimate on MATH is needed; MATH is handled by symmetry) and abbreviate MATH. Using the shorthand MATH, we have MATH . Indeed, in the first line we noted that MATH because MATH. Then we took out the terms MATH as well as MATH, recalling that MATH and that MATH. The last inequality follows by the fact that MATH. Invoking REF, the sum in the exponent is bounded above by MATH, whereby we finally get that MATH.
math-ph/0004014
Since MATH is continuous, there is a ball MATH of radius of order MATH such that MATH. If MATH is so large that MATH, then MATH and the left-hand side of REF is bounded from above by MATH. The proof now proceeds in a different way depending whether MATH or MATH. In the following, the words ``percolation," ``infinite cluster," etc., refer to site-percolation on MATH with parameter MATH. Recall that MATH by our choice of MATH. Let MATH. Then, by equality of MATH and the limit of slab-percolation thresholds, there is a width MATH such that the slab MATH contains almost surely an infinite cluster. Pick a lattice direction and decompose MATH into a disjoint union of translates of MATH. There is MATH such that, for MATH large, at least MATH slabs are intersected by MATH. Then MATH is contained in the event that in none of the slabs intersecting MATH the respective infinite cluster reaches MATH. Let MATH be minimum probability that a site in MATH belongs to an infinite cluster. Combining the preceding inclusions, we have MATH . Now the claim follows by putting MATH. In MATH, suppose without loss of generality that MATH is centered at the origin. Recall that MATH and MATH are MATH-connected if their Euclidean distance is not more than MATH. On the event MATH, the origin is encircled by a MATH-connected circuit of size at least MATH for some MATH, not depending on MATH. Denote by MATH the nearest point of this circuit in the first coordinate direction. Call sites MATH with MATH ``occupied," the other sites are ``vacant." Note that percolation of occupied sites rules out percolation of vacant sites, for example, by the result of CITE. Moreover, using the site-perolation version of the famous MATH result (see for example, CITE), the probability that a given site is contained in a vacant MATH-cluster of size MATH is bounded by MATH, where MATH since MATH. If the ball MATH has diameter at least MATH, then by taking the above circuit for such a cluster we can estimate the probability of its occurrence: MATH for MATH large enough. Here MATH in the sum accounts for the position of the circuit's intersection with the positive part of the first coordinate axis. The minimal size of the circuit is at least MATH, since it has to stay all outside MATH. The claim follows by putting MATH.
math-ph/0004014
Let MATH be the solution to MATH and let MATH be its scaled version: MATH. Define the tilted probability measure MATH . We denote expectation with respect to MATH by MATH. Consider the event MATH . Then MATH can be bounded as MATH . Applying the left inequality in REF, we obtain MATH . Since MATH and MATH is continuous and bounded, we can use our Scaling Assumption and the fact that MATH to turn the sum over MATH into a NAME integral over MATH: MATH where we also used that MATH in this case. In order to finish the proof of the lower bound in REF, we thus need to show that MATH and that MATH . Let us begin with REF. For simplicity, we restrict ourselves to the case when MATH. Then MATH and MATH. Hence, MATH where MATH. Since MATH for any MATH, REF is proved. In order to prove REF, note that MATH . We concentrate on estimating the second term; the first term is handled analogously. By the exponential NAME inequality, we have for any MATH that MATH . Note that MATH (recall REF) as MATH uniformly on compact sets in MATH. Also note that MATH is bounded away from MATH. Choose MATH, where MATH is still to be chosen appropriately. Then the exponent in the third line of REF can be bounded from above by MATH where we replaced MATH by MATH and used the definition relation for MATH. Pick MATH for definiteness. Taking the product over MATH in REF and using that MATH, we obtain for MATH large that MATH where also used that MATH and MATH for some MATH as MATH. By our choice of MATH, REF is clearly satisfied, which finishes the proof in the case MATH.
math-ph/0004014
Suppose MATH has an atom at MATH with mass MATH. Then, noting that MATH are only the sites with MATH, we have MATH . Since MATH and MATH, we have MATH and MATH, whereby REF immediately follows.
math-ph/0004014
Suppose that MATH and MATH. Set MATH and consider the probability measure MATH with density MATH with respect to MATH. Invoking that MATH, we obtain MATH . Now use the Scaling Assumption and the fact that MATH as MATH to extract the term MATH from the exponential on the right-hand side (here we recalled that MATH). Moreover, by an argument similar to REF, the last term on the right-hand side is no smaller than MATH as MATH. To that end we noted that our choice of MATH corresponds to MATH and then we used again that MATH, which follows from the fact that MATH has no atom at zero. This finally finishes the proof of REF .
math-ph/0004014
Fix MATH and MATH with MATH. Recall the notation REF. Let MATH be such that for all MATH and for all MATH . Such a MATH indeed exists, since MATH as MATH and since MATH is uniformly continuous on MATH. This implies that to prove REF it suffices to find an almost-surely finite MATH such that for each MATH there is a MATH for which the event MATH occurs. Indeed, for any MATH with MATH and MATH we have that MATH and MATH, as follows by monotonicity of the maps MATH and MATH and, consequently, MATH by invoking REF. Then REF would follow with the choice MATH. Based on the preceding reduction argument, let MATH for the remainder of the proof. Let MATH. We claim that, to prove REF for MATH, it suffices to show the summability of MATH . Indeed, since MATH we have by REF MATH . Since we assumed MATH, summability of MATH would imply the existence of at least one site MATH (in fact, at least MATH sites) with MATH satisfied. To prove a suitable bound on MATH we invoke NAME 's inequality to find that MATH . As follows from REF, the first term on the right-hand side is summable on MATH. In order to estimate MATH for MATH, let MATH and MATH be two disjoint half spaces in MATH which contain MATH and MATH, respectively, including the outer boundaries. By our choice of MATH, MATH can be chosen such that MATH, and similarly for MATH. We introduce the event MATH that the outer boundary of MATH is connected to infinity by a path in MATH, and the analogous event MATH with MATH and MATH instead of MATH and MATH. By splitting MATH into MATH and MATH (and analogously for MATH) and invoking the independence of MATH and MATH we see that MATH where we recalled REF for the definition of MATH. In order to estimate the last expression, let us observe that MATH where MATH is the event that MATH is in a finite component of MATH which reaches up to MATH. By REF , the probability of the first event is bounded by MATH and, as is well known (see, for example, CITE, proof of REF ), MATH is exponentially small in MATH, which is at least MATH. Since MATH, we have MATH for some MATH. Since MATH for MATH, also the second term is thus summable on MATH, because by REF, MATH. Combining all the preceding reasoning, the proof of REF is finished.
math-ph/0004024
Local exactness of a vertical complex on a coordinate chart MATH follows from a version of the NAME lemma with parameters. We have the the corresponding homotopy operator MATH where MATH. Since MATH is a vector bundle, it is readily observed that the homotopy operator is globally defined, and so is the form MATH.
math-ph/0004024
Bearing in mind the monomorphism REF , let MATH be a closed form whose cohomology class belongs to MATH. We will show that, if MATH is MATH-exact, MATH is exact. The form MATH is represented by the sum REF , and we have MATH . Let MATH where MATH is an exterior form on some finite-order jet manifold MATH. Given a global section MATH of the fibre bundle MATH, let MATH be its MATH-order jet prolongation to a section of the jet bundle MATH. Then it is readily observed that MATH .
math-ph/0004024
Firstly, let us describe the cohomology group MATH of the first row in REF . In this complex, the cocycles are closed forms MATH, while the coboundaries are exact forms MATH where MATH. Because of the isomorphism REF , such a MATH-cocycle is a coboundary in the infinite-order NAME complex and is decomposed into the sum MATH where MATH is a horizontal form such that MATH. It follows that MATH, that is, MATH is a MATH-closed form. At the same time, if MATH is a MATH-exact form MATH, then MATH is a coboundary in MATH. Therefore, we have an epimorphism of cohomology groups MATH . Obviously, its kernel MATH contains the cohomology group MATH. We show that MATH. Let MATH, that is, MATH . Then we have MATH, and MATH in accordance with the decomposition REF . Moreover, being horizontal, MATH that is, it belongs to the MATH-cohomology class of the form MATH on MATH. At last, since all cocycles in MATH are coboundaries in MATH, the morphism MATH sends the cohomology group MATH to zero.
math-ph/0004024
Let MATH be a MATH-closed form, that is, MATH. It belongs to MATH if MATH. In accordance with the relation REF , it follows that MATH, that is, belongs to MATH.
math-ph/0004024
The obstruction for the cohomology group MATH to be trivial are the elements of the overlap MATH, that is, the forms MATH such that MATH, that is, MATH. By virtue of REF , such a form is given by the sum REF which reads MATH . Then, MATH is an exact form.
math/0004002
The assertion follows from REF , which is a part of REF (for reductive MATH). Indeed, reductivity of MATH implies reductivity of MATH CITE. If MATH is not finite, then it contains a non-trivial one-parameter subgroup MATH. For MATH, we have MATH whenever MATH.
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As MATH does not contain MATH, there exists a non-trivial one-parameter subgroup MATH in MATH with finite intersection with MATH. The corresponding extension MATH is spherical iff MATH is spherical in MATH.
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Under these assumptions, there exists a non-trivial one-parameter subgroup of MATH with finite intersection with MATH which normalizes (and even centralizes) MATH.
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Suppose that there exists an affine embedding MATH with infinitely many MATH-orbits. Consider the morphism MATH. It determines an embedding MATH. Let MATH be the integral closure of the subalgebra MATH in the field of rational functions MATH. We have the following commutative diagrams: MATH . The affine variety MATH with a natural MATH-action can be considered as an affine embedding of MATH. The embedding MATH defines a finite (surjective) morphism MATH and therefore, MATH contains infinitely many MATH-orbits. This contradiction completes the proof.
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CASE: By NAME 's theorem, there exists a MATH-module MATH and a vector MATH having REF . Let us denote by MATH the character of MATH at MATH. Since MATH is observable in MATH, every finite-dimensional MATH-module can be embedded in a finite-dimensional MATH-module CITE. In particular, there exists a finite-dimensional MATH-module MATH containing MATH-eigenvectors of character MATH. Choose among them a MATH-eigenvector MATH and put MATH and MATH. REF are satisfied. If REF also holds, then we are done. Otherwise, consider any MATH-module MATH having a vector with stabilizer MATH. Take a MATH-eigenvector MATH with nontrivial character, and replace MATH by MATH and MATH by MATH. Now REF - REF are satisfied. CASE: Since MATH is not spherical in MATH, by a result due to CITE, we may choose MATH in REF so that REF are satisfied. Then we proceed as in REF to obtain the couple MATH. The closure MATH is contained in the image of the NAME embedding MATH and projects MATH-equivariantly onto MATH. This implies REF for MATH. CASE: Let MATH be a fundamental weight of the group MATH. Suppose that MATH acts at the vector MATH constructed above by a character MATH. Since MATH is reductive (and, in particular, is observable), there exists a MATH-module MATH and a MATH-eigenvector MATH with weight MATH CITE. It remains to replace MATH by MATH and MATH by MATH.
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Let MATH be the couple from REF . Denote by MATH the stabilizer MATH of the vector MATH. By REF - REF and since MATH is isomorphic to MATH, MATH is an overgroup of MATH with MATH. By REF, the closure of MATH in MATH is a cone, so by REF does not hold for MATH. REF completes the proof.
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For completeness, we give the proof in the case, where MATH is affine (the only case we need below). It suffices to prove that any MATH-invariant element of MATH is the residue class of a MATH-invariant rational function on MATH. For any MATH, we shall write MATH if MATH. Such ``congruences" are MATH-stable and may be multiplied term by term, as usual numerical congruences. Assume MATH, MATH, MATH, and the residue class of MATH belongs to MATH. Then MATH, and MATH, MATH, that is, MATH. Let MATH be a complementary MATH-submodule to MATH in MATH, and MATH be the projections of MATH on MATH. Then MATH, MATH. By the NAME - NAME theorem, we may choose finitely many MATH, MATH so that MATH is a MATH-eigenfunction in MATH of some weight MATH. Put MATH. Then MATH, whence MATH, MATH. It follows that MATH, hence MATH, because MATH. Thus MATH are MATH-eigenfunctions of the same weight, and MATH has the same residue class in MATH as MATH.
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If MATH is as above, then Knop has shown that the algebra MATH admits a non-trivial MATH-invariant grading, whose homogeneous components are sums of isotypic components of the MATH-module MATH, see CITE and its proof. This grading is constructed as follows. Under the above assumptions, there is a central valuation MATH of MATH such that the respective linear function MATH on MATH lies in MATH, hence MATH vanishes on MATH. In view of MATH, this MATH defines a grading of MATH such that isotypic components MATH are homogeneous of degree MATH. Let MATH be the one-dimensional torus corresponding to this grading. Take any MATH, MATH, MATH. By the NAME - NAME theorem, we may choose finitely many MATH, MATH so that MATH is a MATH-eigenfunction of some weight MATH. Then MATH is a MATH-eigenfunction of the same weight, and MATH. Since MATH, the torus MATH acts on them by the same weight MATH, thence MATH. This shows the inclusion MATH.
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It remains to prove that REF holds for MATH whenever any extension of MATH by a one-dimensional torus is spherical. As MATH is reductive, MATH is reductive, too. If there exists no one-parameter extension of MATH at all, then MATH is finite and MATH is affinely closed by REF . Otherwise MATH. As the spherical case is clear, we may suppose MATH. Let MATH be an affine embedding of MATH. In order to prove that MATH has finitely many MATH-orbits, we may assume that MATH is normal. If MATH contains a proper source, then a one-dimensional torus MATH provided by REF yields a non-spherical extension of MATH. Indeed, if MATH is the preimage of MATH in MATH, then MATH, since MATH. This implies MATH, a contradiction. If MATH contains no proper source, then any proper MATH-stable subvariety MATH is the center of a non-central MATH-invariant valuation MATH. There is an inclusion of residue fields MATH. By REF , MATH is the residue field of the restriction of MATH to MATH, which is the field of rational functions in one variable. As MATH is non-central, MATH, thence MATH is spherical. It follows that MATH has finitely many orbits. (Otherwise, a one-parameter family of MATH-orbits provides a non-spherical MATH-subvariety.)
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The reductive group MATH acts on MATH, which is the field of rational functions on a projective line. If the kernel of this action has positive dimension, then it contains a one-dimensional torus extending MATH to a non-spherical subgroup. Otherwise, either MATH is finite or MATH and each subtorus of MATH has a dense orbit on the projective line. The corollary follows.
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We follow the proof of REF . Put MATH; then MATH. We modify the proof of REF to obtain a MATH-module MATH and a MATH-eigenvector MATH such that MATH, MATH is a finite extension of MATH, and MATH contains infinitely many MATH- (not MATH-) orbits. Arguing as in the proof of REF , we see that the closure MATH of MATH is MATH-stable and has infinitely many MATH-orbits, QED . (Observe that MATH may be not reductive, but REF , required in the proof, does not use the reductivity assumption.) To construct such a couple MATH, it suffices, in the notation of REF , to construct a MATH-module MATH and a vector MATH such that MATH and MATH has infinitely many MATH-orbits. Then we proceed as in REF, replacing MATH by MATH. (Note that the reductivity of MATH is not essential in REF.) It remains to construct a couple MATH. For this purpose, we refine NAME 's construction CITE. By assumption, MATH, hence there exists a character MATH such that for the associated line bundle MATH on MATH, the multiplicity of a certain simple MATH-module MATH in MATH is greater than one CITE. The group MATH acts on MATH and on the isotypic component MATH by MATH-module automorphisms. Take a MATH-module MATH and a vector MATH such that MATH. Let MATH be the closure of MATH in MATH. The natural rational map MATH is MATH-equivariant. Consider a decomposition MATH into irreducible MATH-submodules and fix isomorphisms MATH. Choose a basis MATH of MATH-eigenvectors with weights MATH in MATH, and put MATH. In projective coordinates, MATH . The closure MATH of the graph of MATH in MATH is MATH-stable. We claim that MATH contains infinitely many MATH-orbits. To prove it, take a strictly dominant one-parameter subgroup MATH. If all MATH, then MATH as MATH, because MATH is a positive linear combination of positive roots for all MATH. We may identify MATH with MATH and consider the NAME embedding MATH. Then MATH, where MATH is the dual basis to MATH, and MATH. As the sections MATH are linearly independent on MATH, MATH intersects infinitely many closed disjoint MATH-stable subvarieties MATH, MATH. This proves the claim, because MATH projects MATH-equivariantly onto MATH. Finally, a MATH-module MATH and a vector MATH such that MATH are the desired, because MATH. The proof is complete.
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The NAME algebra of MATH equals MATH, where MATH is the centralizer of MATH in MATH. We have MATH, and MATH. If MATH is semisimple, then MATH, and MATH is finite. REF implies the assertion for this case. Now suppose that MATH is not semisimple. If there exists a non-spherical extension of MATH by a one-dimensional torus MATH, then by REF , there exists a very symmetric affine embedding of MATH with infinitely many MATH-orbits. Finally, suppose that any extension of MATH by a one-dimensional torus in MATH is spherical. Then MATH. As the spherical case is clear, we may assume that MATH. The connected kernel MATH of the action MATH acts on isotypic components of MATH by scalar multiplications. Whence MATH is diagonalizable and central in MATH. By assumption, MATH. Hence MATH is a one-dimensional torus acting on MATH with finite kernel. By REF holds for MATH. The proof is complete.
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We use exactly the same arguments as in the proof of REF replacing an embedding of MATH with infinitely many orbits from CITE by an embedding of MATH-modality MATH constructed in CITE.
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Clearly, MATH. Taking an affine cone over the projective embedding constructed in REF , one obtains an affine embedding of MATH of modality MATH, where MATH is a finite extension of MATH. Using the construction from the proof of REF , we get an affine embedding of MATH of modality MATH. The obvious inequality MATH completes the proof.
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REF follows from REF . To prove REF , we can use REF . If there exists a one-dimensional torus in MATH such that the extension MATH is non-trivial and MATH, then there exists an affine embedding of MATH of modality MATH. Conversely, suppose that MATH is an affine embedding of modality MATH. We need to find a one-dimensional subtorus MATH such that for the extended subgroup MATH we will have MATH. By the definition of modality, there exists a proper MATH-invariant subvariety MATH, such that the codimension of a generic MATH-orbit in MATH is MATH. Therefore, MATH. Consider a MATH-invariant valuation MATH of MATH with the center MATH. For the residue field MATH we have MATH, hence MATH. If the restriction of MATH to MATH is not trivial, then by REF , MATH, a contradiction. Thus MATH is central, and MATH is a source of MATH. A one-dimensional subtorus MATH provided by REF yields the extension of MATH of the same complexity.
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If MATH is not semisimple, then for a central one-dimensional subtorus MATH one has MATH. If MATH is semisimple, then for any one-dimensional subtorus MATH there exists a NAME subgroup MATH which does not contain MATH, and there is a MATH-orbit on MATH of dimension MATH. This implies MATH.
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Given any morphism MATH we construct a morphism MATH as follows: For all MATH, define a map MATH in the obvious way: send MATH with MATH to the morphism MATH that sends MATH to MATH. Now define MATH to be the composite MATH where MATH is the product operation of MATH. This map is easily see to preserve the MATH-action. We must now verify that this preserves compositions. Let MATH and MATH and consider the MATH composition MATH: MATH . We begin by decomposing some maps (using the associativity of MATH) to get MATH . Now we permute two maps to get MATH . The conclusion follows by permuting the second and third map from the top with MATH.
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This is due to REF and the requirement that every morphism preserve the unit map.
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Here MATH is the standard graded module-pullback MATH. It is clear that this kernel is an operad, and a glance at the exact sequence in homology induced by MATH (and the fact that MATH and MATH induce homology isomorphisms in all dimensions and for all degrees - see REF) shows that it is also MATH.
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Suppose MATH is some element. Then we have a commutative diagram MATH which we can expand to a diagram MATH which gives us an element of MATH. To conclude that this defines a morphism of operads, we must show that it respects compositions as in REF, or that the kernel of the map MATH of graded MATH-modules is an ideal with respect to all compositions. Suppose MATH maps to MATH in MATH. Then the image of MATH (in MATH) must have at least one factor that is in MATH - say the MATH. If we form a composition with MATH on another factor - that is, take MATH with MATH - then this factor of MATH is unchanged and appears in the composition. Consequently, the composition has a factor of MATH and maps to MATH in MATH. On the other hand, if we formed the composition MATH the result will have many factors of MATH because the co-endomorphisms in MATH preserve MATH.
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Its inverse sends a morphism MATH to MATH.
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Its inverse sends a morphism MATH to MATH.
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The map MATH fits into an exact sequence MATH where MATH. The first equivariance condition in the definition of an operad (see REF) implies that MATH maps into MATH, where MATH is induced by the composite MATH . It follows that MATH so that the image of the kernel of MATH lies in the kernel of MATH and the MATH is well-defined. The remaining statement follows from the first equivariance condition in REF.
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By REF our free algebra is given by MATH REF implies the existence of chain maps MATH for all MATH and MATH-tuples MATH of positive integers. The map on the left is just transposition of the factors. This induces a morphism of NAME MATH . The associativity conditions imply that compositions are preserved by MATH, and the second statement of REF implies that MATH is MATH-linear, where MATH acts on the right by permuting factors of MATH. The injectivity of MATH follows from the fact that one of the targets of any MATH is MATH, with a MATH-basis of elements of MATH mapping to their image in MATH.
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In this case, we set MATH . As before, we make crucial use of REF . We must proceed more delicately than in REF, though. We will use the approach to operads in REF and define composition operations. A composition MATH (where MATH) implies the existence of a chain-map MATH - this is just the transposition and adjoint. Dualizing gives rise to a map MATH and it is important to realize that this represents MATH distinct maps (corresponding to the different composition operations we may perform). Hence the real target of MATH is MATH - in fact it is MATH . Injectivity of the structure-map follows from its sending MATH to among other things MATH .
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This result is due to NAME. The uniqueness of MATH follows by induction and: CASE: MATH is determined by its values on MATH CASE: the image of the contracting chain-homotopy, MATH, lies in MATH. CASE: the boundary map of MATH is injective on MATH (which implies that there is a unique lift of MATH into the next higher dimension).
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Existence and uniqueness follow from REF, setting MATH and MATH and the fact that, in dimension MATH, so any chain maps between them that commute with augmentation must be equal. In dimension MATH for all MATH. In higher dimensions, set MATH for MATH, MATH and for all MATH. The boundaries of MATH and MATH can be arbitrary elements of MATH and MATH, respectively (one dimension lower), so we must compute the composition of the boundaries via the formulas: MATH and MATH given by the equivariance conditions in REF. Associativity and commutativity follow from the same construction carried out with-MATH and MATH. In this case, we compare the maps MATH and MATH . Both compositions agree in dimension MATH and MATH . Furthermore, the equations for extending the MATH from MATH to all of MATH must agree, since they are given by the composition operations on the operad MATH - see REF. Consequently, MATH in all dimensions. A similar argument (with only one factor, obviously) implies that MATH preserves coproducts. See CITE for an explicit formula for the composition operations of MATH. The compositions given in that paper must agree with the ones proved to exist in REF, because they satisfy the defining conditions and because such compositions are unique.
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We have already proved that MATH is an operad - see REF. To see that the coproduct is an operad morphism, observe that the composition operations of MATH are coalgebra morphisms. This follows from the NAME Theory of Constructions once again, observing that the MATH can be defined by this theory - as the unique map MATH with the properties: CASE: MATH CASE: MATH CASE: MATH is a morphism of MATH-modules (where MATH is equipped with the diagonal MATH-action). CASE: MATH, where MATH, MATH is the contracting cochain in REF, and MATH is the augmentation (so MATH is a contracting cochain of MATH). Having done this, we would only need observe that composing MATH with the MATH preserves these defining conditions.
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For all MATH, let MATH denote the standard MATH-simplex, whose vertices are MATH and whose MATH-faces are MATH, with MATH, MATH. We define MATH to be a free functor on models that are simplices and use the NAME Theory of Constructions to define maps: MATH where MATH is a MATH-simplex with chain complex MATH, and MATH acts on MATH by permuting factors. These maps must make the following diagram commute: MATH and for all MATH and MATH, where MATH and MATH is the map that shuffles the factor MATH to the right of MATH factors of MATH. We define a contracting cochain on MATH via MATH where MATH, MATH, denotes a MATH-dimensional face of MATH. We can define a corresponding contracting homotopy on MATH via MATH, where MATH . Above dimension MATH, MATH is effectively equal to MATH. Now set MATH and MATH. In dimension MATH, we define MATH for all MATH via: MATH . This clearly makes MATH a coalgebra over MATH. Suppose that the MATH are defined below dimension MATH. Then, by acyclic models, MATH is well-defined and satisfies the conclusions of this theorem. We define MATH by induction on MATH, requiring that the following invariant condition be satisfied: MATH - in other words, the leftmost factor must be in MATH. Now we simply set MATH where MATH. The upper line follows from induction on the dimension of MATH and the lower line follows from acyclic models and the induction on MATH. The NAME Theory of Constructions implies that this map is uniquely determined by the invariant condition and the contracting homotopy MATH. It follows that REF must commute since: CASE: any composite of MATH-maps will continue to satisfy the invariant condition. CASE: MATH so that composing a MATH-map with MATH results in a map that still satisfies the invariant condition. CASE: the diagram commutes in lower dimensions (by induction on MATH and acyclic models)
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We form the push out of coalgebras over MATH: MATH which is possible because of REF . This gives us the diagram MATH where MATH and MATH are induced by the inclusions. They are clearly m-coalgebra morphisms.
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We will prove this for MATH - the arguments for MATH and MATH are identical. The claim follows by induction on the number of columns. We show that the following equation (in MATH) is satisfied at column MATH of REF for all MATH: MATH . Every commutative subdiagram of the form: MATH results in an equation MATH and any commutative subdiagram of the form MATH where MATH and MATH are elementary equivalences, results in an equation of the form MATH so that the validity of REF in column MATH implies its validity in column MATH. The conclusion follows. The following converse was stated but not proved by NAME, NAME and NAME in CITE:
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Again, we will prove the result for MATH - the arguments for MATH and MATH are identical. We use the description of localized categories in CITE: they define it via ``generators and relations". The only relations that exist in MATH are:
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Again, we only prove the statement for MATH. This is an immediate consequence of REF. The statement that MATH is an elementary equivalence follows from the fact that, as a chain-complex, MATH, where MATH is acyclic and MATH-torsion free. The equation in MATH follows from the commutativity of REF, which implies that MATH. The invertibility of MATH and MATH (in MATH) implies that we can rewrite this equation as MATH.
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The first statement follows immediately from REF, which implies that we can permute leftward elementary equivalences with the rightward maps and compose - thus simplifying the morphism. The second statement follows from the naturality of the push-outs used in the simplification: we can simplify the upper and lower rows of a hammock in such a way that there is a map of the simplifications.
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This is an immediate consequence of REF.
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The first statement follows from the fact that MATH is full. The second follows from the fact that, in an abelian category, we can always form push-outs. We, consequently, take a diagram of elementary homotopies (like REF) and, whenever we encounter a subdiagram of the form MATH and we form the push out of MATH . It is straightforward to check that this push-out can be inserted into REF . After many cycles of consolidating morphisms going in the same direction and repeating this push-out construction, we end up with a diagram that has the required form. The final claim follows from page REF.
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If MATH is a morphism in MATH, MATH, or MATH, this is clear: the inclusion is a direct summand (as a chain-complex) and a homology equivalence. In the general case (that is, in MATH, MATH, or MATH) MATH is a composite of two elementary equivalences (in opposite directions).
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The only-if part of this is clear. The if part follows immediately from the existence of algebraic mapping cylinders.
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This follows from results in CITE, which characterizes the homotopy category of simplicial sets as a localization of the category of simplicial sets by weak equivalences. REF on page REF shows that a morphism MATH is an isomorphism in the homotopy category if and only if: CASE: MATH is an isomorphism for any set MATH CASE: MATH is an isomorphism for any group MATH CASE: MATH for an local coefficient system MATH of abelian groups on MATH and any MATH . In our case (pointed, simply-connected, REF-reduced simplicial sets), these conditions reduce to MATH inducing homology isomorphisms with MATH coefficients. Since, MATH, where MATH is the morphisms inducing homology isomorphisms, it follows from REF that MATH carries elements of MATH to the saturation of MATH in MATH, hence extending (uniquely) to a functor MATH .
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Consider an elementary homotopy MATH . It induces MATH where the map MATH is an equivalence since MATH in MATH. This factors through to the algebraic mapping cones.
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By abuse of notation, we will denote the algebraic mapping cylinder by MATH (even though MATH is not necessarily a morphism). Let the chain complex of MATH (the unit interval) be given by:
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We construct MATH to have an underlying chain complex that is the algebraic mapping cylinder of MATH. Since MATH is a natural transformation, and since MATH is free on the MATH, it follows that MATH is constructed from MATH. Consequently, REF implies that we can extend the m-structures of MATH and MATH to an m-structure on MATH. These m-structures induce an m-structure on on all of MATH. With this m-structure, the inclusion of MATH and MATH into the ends of this algebraic mapping cylinder are morphisms of m-coalgebras, with the inclusion of MATH an equivalence. The conclusion follows.
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It is only necessary to note that these are functors from the category of ordered pairs of simplicial sets and that the respective functors are free and acyclic on models composed of pairs of simplices.
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The key fact here is that the m-structure of MATH is trivial, thereby eliminating any obstruction to the existence of MATH. The statements about co-commutativity and co-associativity follow immediately from the definition of MATH and the remaining statements are clear.
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This follows from the Duality REF , which states the existence of an operad morphism: MATH . Now we compose the structure map of MATH with this to get an operad morphism MATH . The conclusion follows by noting that, as a non-MATH operad MATH.
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See CITE for a proof. This involves considering MATH and applying the identities that modules and comodules must satisfy.
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REF follow by an inductive application of the boundary formula in REF. The formula for MATH follows by a straightforward but tedious computation.
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Most of the statements follow from the functoriality of MATH. The last statement follows by considering the MATH and passing to the limits: it is not hard to see that MATH for finite MATH. When we pass to the limits, there still exists a map, but it is not necessarily an isomorphism.
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Recall that, in the notation of REF, MATH. The boundary of MATH is MATH, where MATH is given, on MATH by (see REF ) MATH and MATH is given by MATH . The diagram MATH commutes because the composite MATH is of degree MATH, and the diagram MATH is also seen to commute, after we permute MATH with MATH. he conclusion follows.
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We must verify that the identities in REF are satisfied. The statement that MATH is a free group-object implies that MATH by a recursive application of REF. The identity in REF gives us: MATH .
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The proof is very similar to REF (we take the duals of all of the maps there).
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This is very similar to the proof of REF - we have formed the mapping sequence representing the tensor product MATH using REF (regarding MATH as a mapping sequence concentrated in the MATH term) and added a term representing the twisting cochain.
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A lengthy computation (see CITE) shows that equivalence of MATH-structures implies an equation like REF in MATH. The corresponding equation in MATH will hold if there exists a coalgebra, MATH, over MATH such that the structure map MATH is injective. But this follows from REF.
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This follows from REF and the acyclicity of a MATH-operad's components.
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We appeal to REF to get a MATH-module structure on MATH. The conclusion follows from REF, and REF.
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The only thing to be proved is that the maps MATH pull back to the upper row as maps of the MATH. But this is an immediate consequence of:
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In this case, we translate the formula in REF into compositions in MATH: MATH with MATH terms, where MATH for MATH and and, for MATH where MATH represents the inclusion induced by raising all indices The expression for MATH is clearly a direct translation of the equation for MATH in REF into compositions in MATH. The term MATH represents a translation of MATH.
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We know the following two things about the upper rows of REF: CASE: They are independent of MATH CASE: The composites of the maps MATH and MATH, respectively with the corresponding vertical maps satisfy the defining identities (see REF) of a mapping sequence (because every square of the diagrams commutes and because the lower rows are mapping sequences). It suffices, therefore, to find a MATH for which the vertical maps in REF are injective. But the existence of such a MATH follows from REF (which implies that we can use the coalgebra given by REF even though it is concentrated in dimensions MATH). This completes the proof.
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We begin with an observation: Any MATH-structure on MATH induces one on MATH. It follows that we can form MATH and MATH and that there exist canonical morphisms MATH and MATH . It is also not hard to see that these morphisms will be injective if the m-structure morphism of MATH is. CASE: The morphisms MATH and MATH are isomorphisms. This follows from the fact that the maps in stage of the corresponding mapping telescopes are isomorphisms. We conclude that equivalent MATH-structures give rise to isomorphic MATH and MATH. CASE: The images of MATH and MATH in REF are independent of the MATH-structure since they are induced by structure maps of MATH (which only depend on the m-structure of MATH).
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See REF.
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Our first claim is proved in REF. Our second claim follows by a variation on the construction in REF: we replace each term of MATH by MATH in all the subscripts in REF. The conclusion follows from the definition of the MATH - particularly the fact that they were defined exactly like the MATH with a shift in indices.
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The first claim follows from REF. To prove the second, let MATH be the adjoint of MATH in REF and let MATH be the adjoint of MATH defined in REF. We form the tensor product MATH . Now we compose this on the right with the map MATH defining the action of MATH on MATH and with MATH on the left to get a map MATH . The tensor products on the left are all untwisted, but a simple computation shows that the effects of replacing MATH by MATH and MATH by MATH, respectively, exactly cancel out (when mapped by MATH), giving a map MATH that is the adjoint of the map, MATH, that we want.
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That MATH is an operad follows from the fact that MATH is an operad for any comodule over a MATH-operad (by the discussion preceding REF) and REF. That these composition operations commute with the coproduct of MATH follows from: CASE: the definition of the coproduct MATH CASE: the fact that MATH (by REF), CASE: for each term of MATH only a single term of MATH will have a nonzero composition with it. This follows from the fact that MATH unless MATH (in REF).
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This is a straightforward consequence of the fact that the MATH are acyclic in above dimension MATH if MATH is MATH. We can use the identity that a twisting cochain must satisfy - REF - to compute MATH inductively. We set MATH, where MATH is a sequence of nonnegative integers and the direct sum is taken over all such sequences. We begin the induction by requiring that MATH sends MATH to MATH whenever MATH is a sequence with a single nonzero element equal to MATH. The identity for a twisting cochain allows us to extend this to canonical basis elements of MATH in higher dimensions and we extend it to all of MATH by setting MATH .
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To conclude that the pullbacks can be taken, we invoke REF.
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We must show that it maps equivalent objects to equivalent objects. It is easy to see that MATH maps split injections (of MATH) into multiplicative split injections (of MATH) and a simple spectral sequence argument shows that maps induced by homology equivalences are homology equivalences.
math/0004003
Let MATH be a sequence of nonnegative integers and let MATH be a direct summand see REF. Note that MATH contains a summand MATH where MATH. REF implies that the composition operations of MATH coincide with those of MATH. Since the twisted differential of MATH consists of terms MATH with at least one of the MATH see REF, it follows that we can project MATH to MATH - despite the fact that it is in the fiber of the twisted tensor product. An examination of the composition-operations in MATH shows that this projection is compatible with compositions because, restricted to MATH, they are a perturbation of the compositions of MATH - and the perturbations lie in summands MATH with at least one of the MATH. This projection induces the map in the conclusion. The projection MATH is clearly compatible with this morphism of MATH-operads because the action of MATH gets killed off. (Lifting lemma) Let MATH, MATH, and MATH be m-coalgebras over the operad MATH and suppose there exists maps in the category MATH such that MATH in positive dimensions. Compose the structure map of MATH with the projection MATH so that MATH becomes an m-coalgebra over MATH. Then the map MATH, for all MATH, defines a morphism in MATH . The qualifier ``in positive dimensions" was necessary for our statement that MATH: All morphism in MATH induce isomorphisms in dimension MATH which, for any m-coalgebra is equal to MATH. First, note that the map MATH is a chain-map. This follows from the fact that the twisting cochain, MATH vanishes on MATH. So the main thing to be proved is that it preserves m-structures. Let the structure maps of MATH and MATH be MATH . As in the proof of REF, let MATH be a sequence of nonnegative integers and let MATH be a direct summand see REF. Now recall the definition of the action of MATH on MATH in REF: the summand MATH acts on MATH in exactly the same manner as MATH acts on MATH. Consequently, the fact that MATH is a morphism in MATH implies that the action of MATH on MATH is compatible with the action of MATH on MATH. It follows that it will suffice to show that the kernel of MATH acts trivially on MATH, at least when restricted to MATH - or MATH . In the notation of REF, MATH consists of CASE: MATH, where MATH denotes anything above dimension MATH. Our conclusion in this case follows simply by the definition of the m-structure of MATH since the image of MATH consists of elements of the form MATH and the action of MATH on MATH is trivial. CASE: the summands MATH with MATH. In this case, our conclusion follows from the fact that we are really interested in the m-structure of MATH - and the action of MATH on MATH gets killed off by the map MATH. This completes the proof.
math/0004003
There are several cases to consider: CASE: Replace MATH by MATH equivalent to it MATH where MATH and MATH are both elementary equivalences. If we combine this equivalence with the morphism above, we get MATH and we put this into canonical form (see REF) by taking the push out MATH where MATH and MATH are elementary equivalences in MATH. The commutative square in the center implies that MATH and the fact that MATH is an elementary equivalence implies that MATH as objects in MATH. CASE: Now we consider the case where MATH is replaced by an equivalent MATH gives the combined diagram MATH where MATH and MATH are elementary equivalences in MATH. We canonicalize this diagram by taking the push out MATH where all four maps in the center square are elementary equivalences. It is easy to see that MATH which proves our claim CASE: The final case to be considered is that of replacing our original morphism MATH one equivalent to it in the localized category MATH. By REF, it suffices to consider an arbitrary elementary homotopy REF . This will fit into a diagram MATH where MATH and MATH are both elementary equivalences and the morphism MATH may go up or down (without loss of generality, we assume it points down). We claim that the morphism MATH must be a homology equivalence. This follows from the fact that MATH and MATH are elementary equivalences (hence homology equivalences). It is not hard to see that MATH induces a morphism in MATH that is a homology equivalence MATH . Now REF implies that MATH as objects in MATH.
math/0004003
First replace REF by canonical representations in terms of morphisms in the nonlocalized category MATH. We get the somewhat gory diagram MATH where CASE: all maps are morphisms in MATH. CASE: the maps MATH, MATH, MATH, and MATH are elementary equivalences in MATH. In order to put the composites MATH and MATH into canonical form (to make use of the diagram's commutativity) we add the push outs MATH where MATH and MATH are elementary equivalences. The homotopy fiber of MATH is MATH and the homotopy fiber of MATH is MATH. The commutativity of the original REF (as morphisms in MATH) implies (by REF) that these homotopy fibers are equal in MATH. Choose a specific equivalence MATH . The morphism MATH is the composite MATH where CASE: the single vertical map, MATH, is always an elementary equivalence in MATH CASE: all other maps, except for MATH, are morphisms in MATH CASE: MATH is an equivalence in the localized category MATH . This clearly defines a morphism in the localized category, MATH that makes REF. If the horizontal morphisms in our original diagram, REF, were equivalences, then MATH and MATH will be equivalences, and so will our morphism MATH.
math/0004003
Let MATH be inclusion of the fiber and consider the commutative diagram of spaces MATH where: CASE: both rows are homotopy equivalences CASE: MATH is the map in REF CASE: MATH is the projection of the fibration to its base This gives a commutative diagram of m-coalgebras, where all maps are defined in the category MATH and the horizontal maps are homology equivalences: MATH . Now we take canonical acyclic twisted tensor products the copies of MATH in the lower row by MATH and pull them back to the upper row to get a commutative diagram MATH where all maps are morphisms in MATH. As before, both rows are homology equivalences. We will mainly be interested in the upper row. Since the twisting cochain MATH vanishes on the image of MATH (inclusion of the fiber), it follows that MATH gives rise to a chain-map MATH sending MATH to MATH . The lifting REF implies that MATH is a morphism in MATH, that is, it preserves m-structures exactly. The map MATH is a homology equivalence. There are several ways to see this. Since we have established that MATH preserves m-structures, we can forget about them and convert MATH into multiple twisted tensor products, using the twisted NAME theorem in CITE. This will give a commutative diagram of chain-maps (not preserving m-structures) MATH and noting that we can ``rearrange" the twisted tensor products on the bottom term to get MATH and noting that this is a twisted tensor product over an acyclic base, so that inclusion of the fiber (which MATH is) induces homology isomorphisms.
math/0004003
This follows from REF by making MATH equal the basepoint of MATH.
math/0004003
This follows immediately from the equivalence of homotopy theories of pointed, REF-reduced, simplicial sets and pointed simply connected spaces - by the adjoint functors MATH (REF-reduced singular complex) and MATH (topological realization). See CITE for details.
math/0004003
The hypothesis implies that the chain-complexes are chain-homotopy equivalent, hence that the MATH, MATH, have the same homology. This implies that the lowest-dimensional nonvanishing homology groups - say MATH in dimension MATH - are isomorphic. We get a diagram of morphisms in MATH . Where CASE: The maps MATH, MATH, are induced by geometric classifying maps; CASE: horizontal maps are equivalences in MATH; Without loss of generality, we may assume that this diagram commutes in the localized category, MATH. Note that, on the chain-level, this means that the diagram is only homotopy-commutative. This claim follows from: CASE: we can make the diagram commute exactly if we replace MATH and MATH by the associated , REF-reduced singular simplicial sets, MATH. This is because the obstruction to altering the maps MATH vanishes above dimension MATH. CASE: MATH of the singular complexes is canonically equivalent, in MATH, to MATH of the originals simplicial sets, by REF implies that there exists an equivalence in MATH such that the following diagram commutes: MATH . Now, REF implies equivalences in MATH for MATH . We conclude that there is an equivalence MATH where MATH denotes the loop space functor and MATH is the canonical twisting function (defining a fibration as twisted Cartesian product - see CITE). In addition, the commutativity of REF implies that MATH where MATH and MATH are the MATH-invariants of the fibrations MATH and MATH, respectively. Since the MATH are homotopy fibers of the MATH maps for MATH, respectively, we conclude that the second stage of the NAME towers of MATH and MATH are equivalent. A straightforward induction implies that all finite stages of the NAME tower of MATH are equivalent to corresponding finite stages of the NAME tower of MATH. We ultimately get the following commutative diagram in MATH: MATH . Passing to topological realizations gives us a homotopy commutative diagram of spaces MATH where: CASE: the MATH are the NAME towers CASE: MATH are the canonical projections (homotopy equivalences) CASE: MATH is a map inserted into the diagram to make it homotopy-commutative. By REF induces a diagram that commutes in MATH: MATH . Combining REF with REF (and invoking REF) give the following commutative diagram in MATH where MATH, MATH are the elementary equivalences induced by the canonical inclusions. This proves the result.
math/0004003
We prove this result by an inductive argument somewhat different from that used in REF . We build a sequence of fibrations MATH over MATH in such a way that CASE: the morphism MATH lifts to MATH - that is, we have commutative diagrams MATH . For all MATH, MATH will be a fibration over MATH with fiber a suitable NAME space. CASE: The map MATH is MATH-connected in homology. If the morphism MATH were geometric, we would be building its NAME tower. Assuming that this inductive procedure can be carried out, we note that it forms a convergent sequence of fibrations (see REF). This implies that we may pass to the inverse limit and get a commutative diagram MATH where MATH is a morphism of m-coalgebras that is a homology equivalence. Now REF implies that MATH is an equivalence of m-coalgebras, and REF implies that it is topologically realizable. It follows that we get a (geometric) map MATH and the composite of this with the projection MATH is a topological realization of the original map MATH. The commutativity of REF follows from that of REF. It only remains to verify the inductive step: Suppose we are in the MATH iteration of this inductive procedure. Then the mapping cone, MATH is acyclic below dimension MATH. Suppose that MATH. Then we get a long exact sequence in cohomology: MATH . Let MATH be the image of MATH and consider the map MATH classified by MATH. We pull back the contractible fibration MATH over MATH (or form the homotopy fiber of MATH) to get a fibration MATH where, as before, MATH represents the loop space. Claim: The morphism MATH lifts to a morphism MATH in such a way that the following diagram commutes: MATH where MATH is that fibration's projection map. Proof of Claim: We begin by REF to conclude the existence of a commutative diagram: MATH where MATH is an m-coalgebra equivalence. If we pull back this twisted tensor product over the map MATH, we get a trivial twisted tensor product (that is, an untwisted tensor product), because the image of MATH, by the exactness of REF. The Lifting REF implies the existence of a morphism MATH . The composition of this map with MATH in REF is the required map MATH . To see that MATH, note that: CASE: MATH is the pullback of the class in MATH inducing a homology isomorphism MATH (by abuse of notation, we identify MATH with a cochain) or MATH CASE: in the stable range, MATH is nothing but the algebraic mapping cone of the chain-map, MATH, above. But the algebraic mapping cone of MATH clearly has vanishing homology in dimension MATH since MATH induces homology isomorphisms. The conclusion follows by obstruction theory.